Problem 683

The surface area of a rectangular prism is given by the formula $A=2\ell w+2\ell h+2wh$. Thus, the rectangular prism with length 2 cm, width 5 cm, and height 9 cm has surface area $2(2)(5)+2(2)(9)+2(5)(9)=20+36+90=146$ cm$^2$. The volume of a rectangular prism is given by the formula $V=\ell wh$. If the rectangular prism has a square base, then $\ell =w$ and so $V=\ell ^2h$. Substituting $V=160$ cm$^3$ and $h=10$ cm, we get $160=\ell ^2(10)$ or $\ell ^2=16$, and so $\ell =4$ cm (since $\ell >0$). Therefore, the side length of the square base of a rectangular prism with height 10 cm and volume 160 cm$^3$ is 4 cm. If a rectangular prism has a square base, then $\ell=w$. Since the area of the base is 36 cm$^2$, then $36=\ell \cdot w=\ell^2$, and so $\ell =w=\sqrt{36}=6$ cm (since $\ell>0$). If the surface area of this prism is 240 cm$^2$, then substituting, we get $240=2(6)(6)+2(6)h+2(6)h$ or $240=72+24h$, and so $h=\dfrac{240-72}{24}=7$ cm. Thus, the volume of the prism is $\ell wh=(6)(6)(7)=252$ cm$^3$. Substituting into the formula for volume, we get $x=k(2k)(3k)$ or $x=6k^3$. Substituting into the formula for surface area, we get $x=2(k)(2k)+2(k)(3k)+2(2k)(3k)$ or $x=4k^2+6k^2+12k^2=22k^2$. Equating the two expressions that are each equal to $x$ and solving, we get $$\begin{aligned} 6k^3& = 22k^2\\ 6k^3-22k^2& = 0\\ 2k^2(3k-11)& = 0\end{aligned}$$ Since $k>0$, then $3k-11=0$ and so $k=\dfrac{11}{3}$.